Correct usage of arrayfun/bsxfun in Matlab - simple example

Question

Assume we have a 1-d Matrix, with random length:

M = [102,4,12,6,8,3,4,65,23,43,111,4]

Moreover I have a vector with values that are linked to the index of M:

V = [1,5]

What i want is a simple code of:

counter = 1;
NewM = zeros(length(V)*3,1);
for i = 1:length(V)
    NewM(counter:(counter+2)) = M(V(i):(V(i)+2))
    counter = counter+3;
end

So, the result will be

NewM = [102,4,12,8,3,4]

In other words, I want the V to V+2 values from M in a new array. I'm pretty sure this can be done easier, but I'm struggeling with how to implement it in arrayfun /bsxfun...

bsxfun(@(x,y) x(y:(y+2)),M,V)

Never throw `bsxfun` (makes things faster!) and `arrayfun`/`cellfun` (just makes them easier...) into the same pot, the latter internally uses loops and you should always preferably use `bsxfun` if possible. — Robert Seifert, Feb 29 '16 at 11:33

O'Neil · Accepted Answer · 2016-02-29T13:31:12.843

3

With bsxfun it's about getting the V_i on one dimension and the +0, +1 +2 on the other:

M = [102,4,12,6,8,3,4,65,23,43,111,4];
V = [1,5];
NewM = M( bsxfun(@plus, V(:), 0:2) )

NewM =
    102     4    12       
      8     3     4

Or if you want a line:

NewM = reshape(NewM.', 1, [])

NewM =
    102     4    12     8     3     4

edited Feb 29 '16 at 13:31

answered Feb 29 '16 at 10:03

O'Neil

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though i should avoid this comment according to stackoverflow: thanks alot! – Jonas Feb 29 '16 at 10:05
1

@Jonas I could add to this answer that the `reshape` can wrap the `bsxfun` to reduce the `NewM` rows vector computation to a single line: `NewM = M(reshape(bsxfun(@plus, V(:), 0:2),1,[]))` – dfrib Feb 29 '16 at 10:50

score 2 · Answer 2 · answered Feb 29 '16 at 09:25

Using arrayfun (note that M is used as an "external" matrix entity in this scope, whereas the arrayfun anonymous function parameter x correspond to the elements in V)

NewM = cell2mat(arrayfun(@(x) M(x:x+2), V, 'UniformOutput', false));

With result

NewM =

   102     4    12     8     3     4

Dan · Answer 3 · 2016-02-29T09:37:53.663

One fully vectorized solution is to expand V to create the correct indexing vector. In your case:

[1,2,3,5,6,7]

You can expand V to replicate each of it's elements n times and then add a repeating vector of 0:(n-1):

n = 3;
idx = kron(V, ones(1,n)) + mod(0:numel(V)*n-1,n)

So here kron(V, ones(1,n)) will return [1,1,1,5,5,5] and mod(0:numel(V)*n-1,n) will return [0,1,2,0,1,2] which add up to the required indexing vector [1,2,3,5,6,7]

and now it's just

M(idx)

Note a slightly faster alternative to kron is reshape(repmat(V,n,1),1,[])

Correct usage of arrayfun/bsxfun in Matlab - simple example

3 Answers3